Periodic Motion Problem, how to tackle this beast?

Question

anonymous · Answer

Two identical, thin rods, each with mass m and length L, are joined at right angles to form an L-shaped object. This object is balanced on top of a sharp edge. If the L-shaped object is deflected slightly, it oscillates.

Find the frequency of oscillation.

anonymous · Answer

I realize that the frequency is$$f=(1/2\pi)\sqrt{mgd/I}$$ where d is the distance from the center of gravity. Any help will be appreciated. Thanks

ujjwal · Answer

What is $d$ and $I$? you mention $d$ is distance from center of gravity.. But from center of gravity to where?

anonymous · Answer

Thanks for replying

d = distance from pivot to center of gravity (CG). 
I for a rod with pivot at one end:  I = 1/3ML^2

the center of mass of each rod is located at L/2 from the pivot. Therefore, the center of gravity of the two rods combined is at the center of the horizontal line that connects the two center of mass points. Using trig, the distance between the pivot and CG is L/2sqrt(2). So we have:

I=(2/3)ML^2 (the sum of the two I's)
d=(L/2sqrt(2)) 

so plugin these values into the eqn. for frequency:

$$f=(1/2\pi)\sqrt{mgd/I}$$ I get: $$f=(1/2\pi)\sqrt{3g/4\sqrt{2}L} = (1/4\pi)\sqrt{3g/\sqrt{2}L}$$ The correct answer is actually: $$f = (1/4\pi)\sqrt{6g/\sqrt{2}L}$$

So it looks like I need to either multiply the equation for d by 2, or multiply the eqn for I by 1/2 - which I don't see why I need to do either. Any ideas?

anonymous · Answer

Jevus fricking Crust .... I finally got this. The issue is in the eqn. for the moment of inertia:
Moment of inertia for one rod is: $$I_{1}=I _{cm}+md ^{2}= (1\12)mL^{2} + m(L/2\sqrt{2})^{2} = (5/24)mL^{2}$$ so I for 2 rods with axis about the center of gravity: $$I_{2}=2I_{1}=(5/12)mL^{2}$$ Now for I with an axis about the pivot point: $$I=I_{2}+2md^{2}=(5/12)mL^{2}+2(m(L/2\sqrt{2})^{2})=(2/3)mL^{2}$$ Plug these into the equation for frequency:$$f=(1/2\pi)\sqrt{2mgd/I}=(1/4\pi)\sqrt{6g/\sqrt{2}L} $$